A cylindrical capacitor has two co-axial cylinders of length 15 cm and radil 1.5 and 1.4 cm.The outer cylinder is earthed and inner cylinder is given a charge of `3.5 muC`. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e., bending of field lines at the ends.)

Explanation

The capacitance of a cylindrical capacitor is given by
`C=(2pi epsilon_(0)LK)/(2.3026"log"_(10)(b)/(a))`
Or, `C=(K*4pi epsilon_(0))/(2) (L)/(2.3026xx "log"_(10)(b)/(a))`, where a and b are outer and inner radii of the cylinders and L is the length of the cylinders respectively.
As, `(b)/(a) = (1.5 xx 10^(-2))/(1.4xx10^(-2))=(15)/(14)`
`rArr C=(1)/(2xx9xx10^(9))xx(15xx10^(-2))/(2.3026xx "log"_(10)(15)/(14))`
`=(15xx10^(-2))/(2xx9xx10^(9)xx2.3026xx0.03)`
`C=1.21xx10^(-10)F`
As the outer cylinder is earthed, the potential of the outer cylinder will be zero. And the potential of inner cylinder will be equal to the potential difference between the cylinders.
`V=(Q)/(C )=(3.5xx10^(-6)C)/(1.21xx10^(-10)F)=2.89xx10^(4)V`